1
$\begingroup$

I am using urca package of R for Johansen Cointegration test in 2 stocks datas( A and B.

My question is very elementar, but have cause some problems for me. How I interpret the critical values, for exemple, H1 <- ca.jo(ll,type = "eigen", ecdet= "const", K = 4,spec = "longrun") produce:

Values of teststatistic and critical values of test:

      test 10pct  5pct  1pct
r <= 1 |  6.39  7.52  9.24 12.97
r = 0  | 11.62 13.75 15.67 20.20

1 - In this case, for 10pct I have 11.62 < 13.75, then I can accept the hipotesis that A and B is not co-intregrate ? (whatever if for r <= 1 test is smaller that critical value ?).

2 - If for result in Trace statistic I find that A and B is cointegrate and for eigenvalue test A and B is not ? what does that mean it ? I reject the result of Trace Statistics in this case and admit A and Bnot cointegrate ?

$\endgroup$
  • $\begingroup$ Hi @FlàvioCorinthians and welcome to quant.SE! Could you improve the question format? For instance, you could add the label to the test (c-value, p-value,..). Moreover, I suggest you to browse in quant.SE, since there are a lot of questions about cointegration test interpretation (see, for example, quant.stackexchange.com/questions/2076/…) $\endgroup$ – Quantopik Jul 13 '15 at 9:54
1
$\begingroup$

Johansen test estimates the rank (r) of given matrix of time series with confidence level. In your example you have 2 time series, therefore Johansen tests null hypothesis of r=0 < (no cointegration at all), r<1 (till n-1, where n=2 in your example). If r<=1 test value (6.39) was greater than a confidence level's value (say 10%: 7.52), we would assume there is a cointegration of r time series (in this case r<=1). But as you see, none of your test values are greater than than critical values at r<0 and r<=1, therefore there is no cointegration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.